How to grep only if pattern1 and pattern2 matches in consecutive lines

user_2441241 0
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I have a file like below:

city-italy
good food
bad climate
-
city-india
bad food
normal climate
-
city-brussel
normal dressing
stylish cookings
good food
-

Question - I want to grep city and food, for which "food" is "bad".

For example - for the above question, i need a grep command to get a answer like below

city-india
bad food

Please help me like, how i will get pattern 1 and pattern 2 grepped only if both succeeds parallely.

i mean both pattern should match and it should grep in the following line.

grep
6 Answers

You can do it with pipes - grep -A1 city <filename> | grep -B1 "bad food" or cat filename | grep -A1 city | grep -B1 "bad food" (or any other stream source for the pipe)

user_25963750

If the city name is guaranteed to come before the food quality (any other info in between is allowed):

sed -n -e '/^city/h' -e '/bad food/{x;G;p}' input

Which keeps the name of each city in the hold buffer and prints the last city name when matches bad food.

user_3909130

If the order is ensured, you can use directly the command grep with OR:

grep -e "city" -e "food" FILE_INPUT

Then hopefully the city will follow by its food feature at following.

The result looks like:

city-italy
good food
city-india
bad food
city-brussel
good food

You can change your pattern to get a more filtered result.

user_14534870

To get city with bad food using gnu awk (due to RS)

awk '/bad food/ {print RS $1}' RS="city" file
city-india
user_23418470

another awk line:

kent$  awk 'BEGIN{FS=OFS="\n";RS="-"FS}/bad food/{print $1,$2}' file
city-india
bad food
user_1648350
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